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Mathematics 9 Online
OpenStudy (anonymous):

Question 15 (Multiple Choice Worth 5 points) [9.05] Solve 3x2 + 5x + 7 = 0. Round solutions to the nearest hundredth. No real solutions x ≈ −6.34 and x ≈ 1.34 x ≈ −0.45 and x ≈ 2.11 x ≈ −2.11 and x ≈ 0.45

OpenStudy (rob1525):

hi felicia123...at it again hu? lol! Okay, so u can use the pathagoriean therom again to find the x intercepts...remember its\[(-b + or - \sqrt{b^2-4*a*c})/2a\]

OpenStudy (rob1525):

\[x1= (b + \sqrt{b^2-4*a*c})/2a\] and \[x2= (b - \sqrt{b^2-4*a*c})/2a\]

OpenStudy (rob1525):

remember that a quadratic has the form a^2+b+c=0

OpenStudy (anonymous):

can you just tell me how to do this in the cal

OpenStudy (rob1525):

okay a=3, b=5 and c=7.... :)

OpenStudy (anonymous):

Is it A

OpenStudy (rob1525):

idk

OpenStudy (anonymous):

i think its a because of the 0

OpenStudy (rob1525):

\[x1= (5 + \sqrt{5^2-4*3*7})/2a\]

OpenStudy (anonymous):

it be 3x ^2-5x+7 =0

OpenStudy (rob1525):

so b= -5

OpenStudy (rob1525):

put -5 for b in the formula

OpenStudy (anonymous):

-5

OpenStudy (rob1525):

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